Difference in preprocessor output - c++

Question

I was trying to check some single line macros which have 2 pre-processor directives.
#define REPLACE { \
#if EXT == 42 \
#warning "Got 42" \
#endif \
}
int main(void){
REPLACE;
return 0;
}
The pre-processor parses this fine yielding:
$g++ -E includetest.cpp
# 1 "includetest.cpp"
# 1 "<built-in>"
# 1 "<command-line>"
# 1 "/usr/include/stdc-predef.h" 1 3 4
# 1 "<command-line>" 2
# 1 "includetest.cpp"
int main(void){
{ #if EXT == 42 #warning "Got 42" #endif };
return 0;
}
which of course is illegal code, since just macro substitution occurs and the ifdef lookalike macro isn't processed again even though it looks like one.
Now if I slightly alter the macro to look something like
#define REPLACE(a) { a + 2 ; \
#if EXT == 42 \
#warning "Got 42" \
#endif \
}
int main(void){
REPLACE(0);
return 0;
}
Which yields this pre-processor error:
$g++ -E includetest.cpp
# 1 "includetest.cpp"
# 1 "<built-in>"
# 1 "<command-line>"
# 1 "/usr/include/stdc-predef.h" 1 3 4
# 1 "<command-line>" 2
# 1 "includetest.cpp"
includetest.cpp:1:18: error: '#' is not followed by a macro parameter
#define REPLACE(a) { a + 2 ; \
^
int main(void){
REPLACE(0);
return 0;
}
Why does this error come up ? Of course, this won't compile but I want to know why addition of a parameter ensues in a parsing error from pre-processor ?
People would say " You can't nest another directive in another", but in the first case too they are nested, why doesn't pre-processor error out then ? Or is that responsibility delegated to the compiler ?
EDIT: I am not trying to achieve any functionality per se, this is just an exercise (in futility?) to understand the pre-processor.

Answer 1

Only inside a function macro does a # have special meaning ([cpp.stringize]p1). The standard says that a # needs to be followed by a function argument, which is not happening in your second case (if, warning and endif are not parameters).
Your first case is valid (you can actually have directives inside the replacement list of object macros) exactly because there # doesn't have any special meaning.